Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 27 - Early Quantum Theory and Models of the Atom - Problems - Page 800: 32

Answer

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Work Step by Step

We find the Compton wavelength shift for a photon scattered from an electron, using Eq. 27–7. $$\lambda ‘=\lambda + \frac{h}{m_e c}(1-cos \phi)$$ $$\Delta \lambda = \frac{h}{m_e c}(1-cos \phi)$$ Calculate the Compton wavelength of a free electron. $$\frac{h}{m_e c} = \frac{6.626\times10^{-34}J\cdot s}{(9.11\times10^{-31}kg) (3.00\times10^8 m/s) }=2.424\times10^{-12}m$$ Finally, we obtain this relation. $$\Delta \lambda = (2.424\times10^{-3}nm)(1-cos \phi)$$ a. Apply the equation for a scattered angle of $45^{\circ}$. $$\Delta \lambda = (2.424\times10^{-3}nm)(1-cos 45^{\circ})=7.1\times10^{-4}nm$$ b. Apply the equation for a scattered angle of $90^{\circ}$. $$\Delta \lambda = (2.424\times10^{-3}nm)(1-cos 90^{\circ})=2.4\times10^{-3}nm$$ c. Apply the equation for a scattered angle of $180^{\circ}$. $$\Delta \lambda = (2.424\times10^{-3}nm)(1-cos 180^{\circ})=4.8\times10^{-3}nm$$
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